International GCSE Maths · Pearson Edexcel 4MA1 · Unit 6
Statistics and probability
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1IGME-063-00-01Averages, range and statistical diagramsPaper FEasy6 marks
Ten customers at a bank recorded how long, in minutes, they waited to be served. 12, 7, 15, 9, 12, 18, 10, 12, 6, 14
(a)
Write down the mode.
(1)
(b)
Find the median.
(2)
(c)
Work out the range.
(1)
(d)
Work out the mean waiting time.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 for 12 minutes
✓ 1
Part (b)
M1 for ordering the values: 6, 7, 9, 10, 12, 12, 12, 14, 15, 18, or for identifying the 5th and 6th values
✓ 1
A1 for 12 minutes
✓ 1
Part (c)
B1 for 18 − 6 = 12 minutes
✓ 1
Part (d)
M1 for a total of 115, then 115 ÷ 10
✓ 1
A1 for 11.5 minutes
✓ 1
Exam tip: Always put the data in order before finding the median. With an even number of values, the median is halfway between the two middle values.
2IGME-063-00-02Averages, range and statistical diagramsPaper FEasy5 marks
A football team played 24 matches. The number of goals the team scored in each match is shown below. Goals scored: 0, 1, 2, 3, 4 Frequency: 5, 8, 6, 3, 2
(a)
Write down the mode.
(1)
(b)
Write down the range.
(1)
(c)
Work out the mean number of goals scored per match. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 for 1 goal
✓ 1
Part (b)
B1 for 4 − 0 = 4 goals
✓ 1
Part (c)
M1 for at least three correct products of goals × frequency: 0, 8, 12, 9, 8
✓ 1
M1 for 37 ÷ 24
✓ 1
A1 for 1.54 goals
✓ 1
Common error: Do not divide by 5, the number of rows. Divide the total number of goals, 37, by the total frequency, 24.
3IGME-063-00-03Averages, range and statistical diagramsPaper FEasy4 marks
At Hillcrest School, 180 students were asked to name their favourite sport. Football: 70 students, Tennis: 35 students, Swimming: 45 students, Other: 30 students. The results are to be shown in a pie chart.
(a)
Work out the angle of the sector for Tennis.
(2)
(b)
At Riverside School, the students' answers are also shown in a pie chart. The sector for Hockey has an angle of 48° and represents 32 students. Work out the total number of students in the Riverside pie chart.
(2)
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Marking point
Mark
Notes
Part (a)
M1 for 360 ÷ 180 = 2° per student, or 35/180 × 360
✓ 1
A1 for 70°
✓ 1
Part (b)
M1 for 32 ÷ 48 × 360, or for 48 ÷ 32 = 1.5° per student
✓ 1
A1 for 240 students
✓ 1
Exam tip: Find the angle for one person first. All the angles in a pie chart must add up to 360°, which is a useful check.
4IGME-063-00-04Averages, range and statistical diagramsPaper FEasy4 marks
The stem-and-leaf diagram gives the scores of 15 students in a quiz. 2 | 5 8 9 3 | 1 3 3 7 4 | 0 2 4 4 4 8 5 | 3 6 Key: 3 | 1 means 31 points
(a)
Write down the mode.
(1)
(b)
Find the median.
(2)
(c)
Work out the range.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 for 44 points
✓ 1
Part (b)
M1 for identifying the 8th value, i.e. (15 + 1) ÷ 2
✓ 1
A1 for 40 points
✓ 1
Part (c)
B1 for 56 − 25 = 31 points
✓ 1
Common error: The median is the 8th value, not 8. Count along the leaves in order and use the key to write the full value.
5IGME-063-00-05Averages, range and statistical diagramsPaper FEasy3 marks
A basketball player scored a mean of 8 points per game in her first 6 games of the season. After her 7th game, her mean was 9 points per game.
(a)
Work out the number of points she scored in her 7th game.
(3)
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Marking point
Mark
Notes
Part (a)
M1 for 6 × 8 = 48
✓ 1
M1 for 7 × 9 = 63
✓ 1
A1 for 15 points
✓ 1
Exam tip: When a mean changes, work with totals: total = mean × number of values. The difference between the totals gives the new value.
6IGME-063-01-01Probability of single and combined eventsPaper FEasy4 marks
A four-coloured spinner is biased. The table gives some of the probabilities of the spinner landing on each colour.
Red: 0.15 Blue: 0.4 Green: x Yellow: x
The probability of landing on green is the same as the probability of landing on yellow.
(a)
Work out the value of x.
(2)
(b)
Ravi spins the spinner 200 times. Work out an estimate for the number of times the spinner lands on yellow.
(2)
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Mark
Notes
Part (a)
M1: 0.15 + 0.4 + x + x = 1, or 2x = 0.45
✓ 1
A1: x = 0.225
✓ 1
Part (b)
M1: 200 × 0.225
✓ 1
A1: 45
✓ 1
Exam tip: The probabilities of all the possible outcomes always add up to 1. Start by writing that equation.
7IGME-063-01-02Probability of single and combined eventsPaper FEasy4 marks
A bag contains 7 red counters, 5 blue counters and 8 white counters only. A counter is taken at random from the bag.
(a)
Write down the probability that the counter is blue.
(1)
(b)
Work out the probability that the counter is not white.
(1)
(c)
The counter is put back. Then 4 more blue counters are added to the bag. A counter is taken at random. Work out the probability that it is blue.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1: \(\frac{5}{20}\) or \(\frac{1}{4}\) or equivalent
✓ 1
Part (b)
B1: \(\frac{12}{20}\) or \(\frac{3}{5}\) or equivalent
✓ 1
Part (c)
M1: 9 blue counters out of a total of 24
✓ 1
A1: \(\frac{9}{24}\) or \(\frac{3}{8}\) or 0.375
✓ 1
Common error: When counters are added, both the number of blue counters and the total number of counters change. Update the denominator as well.
8IGME-063-01-04Probability of single and combined eventsPaper FEasy3 marks
Sam throws a drawing pin 250 times. It lands point up 95 times.
(a)
Work out the relative frequency of the drawing pin landing point up.
(1)
(b)
Sam throws the same drawing pin another 600 times. Work out an estimate for the number of times it will land point up.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1: 0.38 or \(\frac{95}{250}\) or equivalent
✓ 1
Part (b)
M1: 600 × 0.38 or 600 × \(\frac{95}{250}\)
✓ 1
A1: 228
✓ 1
Common error: Do not just add 95 to the number of throws. Multiply the relative frequency by the new number of trials.
9IGME-063-03-03Conditional probability and tree diagramsPaper FEasy6 marks
The probability that Lena's morning train is late on any day is 0.15. Whether the train is late on one day is independent of any other day. Lena catches the train on Monday and on Tuesday.
(a)
Write down the probability that the train is not late on Monday.
(1)
(b)
Work out the probability that the train is late on exactly one of the two days.
(3)
(c)
Work out the probability that the train is late on at least one of the two days.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 for 0.85
✓ 1
Part (b)
M1 for 0.15 × 0.85 or 0.85 × 0.15
✓ 1
M1 for 0.15 × 0.85 + 0.85 × 0.15 or 2 × 0.15 × 0.85
✓ 1
A1 for 0.255
✓ 1
Part (c)
M1 for 1 − 0.85 × 0.85, or 0.15 × 0.15 + 0.255, or 0.15 + 0.85 × 0.15
✓ 1
A1 for 0.2775
✓ 1
Exam tip: For "at least one", subtracting the probability of "none" from 1 is usually the quickest method.
10IGME-063-03-01Conditional probability and tree diagramsPaper FEasy5 marks
A bag contains 5 red counters and 3 blue counters only. Priya takes a counter at random from the bag, records its colour and puts it back in the bag. She then takes a second counter at random from the bag and records its colour.
(a)
Work out the probability that both counters are red.
(2)
(b)
Work out the probability that the two counters are different colours.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 for 5/8 × 5/8
✓ 1
A1 for 25/64 oe
✓ 1
Part (b)
M1 for 5/8 × 3/8 or 3/8 × 5/8 (15/64), or for 3/8 × 3/8 (9/64) as part of a complement method
✓ 1
M1 for 5/8 × 3/8 + 3/8 × 5/8, or for 1 − 25/64 − 9/64
✓ 1
A1 for 30/64 = 15/32 oe, e.g. 0.46875 or 0.469
✓ 1
Common error: "Different colours" can happen in two orders: red then blue, or blue then red. Using only one of these products halves the answer.
11IGME-063-01-03Probability of single and combined eventsPaper FEasy4 marks
Leila has a fair five-sided spinner numbered 1, 2, 3, 4 and 5, and a fair coin. She spins the spinner once and flips the coin once.
(a)
List all the possible outcomes. One has been done for you: (1, Heads).
(2)
(b)
Work out the probability that the coin shows tails and the spinner shows a number greater than 3.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1: at least 4 further correct outcomes listed, in addition to the given (1, Heads)
✓ 1
B1: all 10 outcomes correct, i.e. (1,H) to (5,H) and (1,T) to (5,T), with no extras or repeats
✓ 1
Part (b)
M1: identifies (4, Tails) and (5, Tails) as 2 of the 10 equally likely outcomes, or uses \(\frac{1}{2} \times \frac{2}{5}\)
✓ 1
A1: \(\frac{2}{10}\) or \(\frac{1}{5}\) or 0.2
✓ 1
Exam tip: List outcomes systematically, for example all the Heads first in number order, then all the Tails, so that none are missed.
12IGME-063-00-06Averages, range and statistical diagramsPaper FMedium6 marks
40 people each completed a puzzle. The time, t minutes, that each person took is summarised below. 0 < t ≤ 10: 4 10 < t ≤ 20: 9 20 < t ≤ 30: 15 30 < t ≤ 40: 8 40 < t ≤ 50: 4
(a)
Write down the modal class.
(1)
(b)
Write down the class interval that contains the median.
(1)
(c)
Work out an estimate for the mean time taken.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 for 20 < t ≤ 30
✓ 1
Part (b)
B1 for 20 < t ≤ 30, since the cumulative totals 4, 13, 28 show the 20th and 21st values lie in this class
✓ 1
Part (c)
M1 for using midpoints 5, 15, 25, 35, 45
✓ 1
M1 for at least three correct products: 20, 135, 375, 280, 180
✓ 1
M1 for 990 ÷ 40
✓ 1
A1 for 24.75 minutes (accept 24.8)
✓ 1
Common error: Multiply each frequency by the midpoint of the class, not by the upper class boundary. Using the end points makes the estimate too large.
13IGME-063-00-07Averages, range and statistical diagramsPaper FMedium5 marks
Ravi recorded the waiting times, in minutes, of 7 customers at Coffee Shop A. 3.1, 4.5, 2.8, 5.0, 3.9, 4.2, 3.6 For Coffee Shop B, the median waiting time was 3.4 minutes and the range was 4.1 minutes.
(a)
Find the median waiting time at Coffee Shop A.
(2)
(b)
Work out the range of the waiting times at Coffee Shop A.
(1)
(c)
Compare the waiting times at the two coffee shops.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 for ordering: 2.8, 3.1, 3.6, 3.9, 4.2, 4.5, 5.0
✓ 1
A1 for 3.9 minutes
✓ 1
Part (b)
B1 for 5.0 − 2.8 = 2.2 minutes
✓ 1
Part (c)
B1 for a comparison of medians in context, e.g. on average customers waited longer at Shop A because its median is higher
✓ 1
B1 for a comparison of ranges in context, e.g. waiting times at Shop A were more consistent because its range is smaller
✓ 1
Exam tip: A full comparison makes one statement about an average and one about spread, and each statement should refer to the context (waiting times), not just to the numbers.
14IGME-063-00-08Averages, range and statistical diagramsPaper FMedium3 marks
Some students were asked how many pets they own. Number of pets: 0, 1, 2, 3 Frequency: 7, x, 9, 4 The mean number of pets is 1.4.
(a)
Work out the value of x.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 for total number of pets = x + 18 + 12 = x + 30, with total frequency 20 + x
✓ 1
M1 for (x + 30) ÷ (20 + x) = 1.4, leading to x + 30 = 28 + 1.4x
✓ 1
A1 for x = 5
✓ 1
Common error: The unknown frequency x appears in both the total number of pets and the total number of students. Leaving it out of the denominator gives the wrong equation.
15IGME-063-00-11Averages, range and statistical diagramsPaper FMedium3 marks
Class A has 18 students. Their mean test mark is 64. Class B has 12 students. Their mean test mark is 71.
(a)
Work out the mean test mark of all 30 students.
(3)
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Mark
Notes
Part (a)
M1 for 18 × 64 = 1152 or 12 × 71 = 852
✓ 1
M1 for (1152 + 852) ÷ 30
✓ 1
A1 for 66.8
✓ 1
Common error: Do not just average the two means, which gives 67.5. The classes are different sizes, so combine the totals first.
16IGME-063-01-07Probability of single and combined eventsPaper FMedium4 marks
Two fair six-sided dice, each numbered 1 to 6, are rolled.
(a)
Work out the probability that the sum of the two numbers is 9.
(2)
(b)
Work out the probability that the product of the two numbers is a multiple of 6.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1: identifies the 4 outcomes (3,6), (4,5), (5,4), (6,3) out of 36
✓ 1
A1: \(\frac{4}{36}\) or \(\frac{1}{9}\)
✓ 1
Part (b)
M1: 15 favourable outcomes out of 36 identified
✓ 1
A1: \(\frac{15}{36}\) or \(\frac{5}{12}\)
✓ 1
Exam tip: Write out a 6 by 6 sample space grid of products, then circle the multiples of 6 so you can count them accurately.
17IGME-063-01-08Probability of single and combined eventsPaper FMedium5 marks
There are 120 students in a year group. 52 are boys and the rest are girls. 18 of the boys wear glasses. In total, 85 students do not wear glasses.
(a)
A student is chosen at random. Work out the probability that the student is a girl who wears glasses.
(3)
(b)
A girl is chosen at random. Work out the probability that she wears glasses.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: 120 − 85 = 35 students wear glasses
✓ 1
M1: 35 − 18 = 17 girls wear glasses
✓ 1
A1: \(\frac{17}{120}\)
✓ 1
Part (b)
M1: 120 − 52 = 68 girls, and \(\frac{17}{68}\)
✓ 1
A1: \(\frac{1}{4}\) or 0.25
✓ 1
Exam tip: Draw a two-way table with totals. When the choice is made only from the girls, the denominator is the number of girls.
18IGME-063-00-13Averages, range and statistical diagramsPaper FHard7 marks
Five positive whole numbers have a mode of 4 a median of 6 a mean of 7 a range of 10.
(a)
Find the five numbers.
(4)
(b)
A sixth whole number is added to the list. The mean of the six numbers is 8. Find the sixth number and the median of the six numbers.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 for 6 as the middle (3rd) number
✓ 1
B1 for the two smallest numbers being 4 and 4
✓ 1
M1 for total 5 × 7 = 35, with largest number 4 + 10 = 14
✓ 1
A1 for 4, 4, 6, 7, 14
✓ 1
Part (b)
M1 for 6 × 8 = 48, then 48 − 35
✓ 1
A1 for the sixth number 13
✓ 1
B1 for the median (6 + 7) ÷ 2 = 6.5 from the list 4, 4, 6, 7, 13, 14
✓ 1
Exam tip: Write five blank spaces in order and fill them in one fact at a time. Then check that your list satisfies every condition, including that the mode is the only one.
19IGME-063-03-02Conditional probability and tree diagramsPaper HEasy4 marks
A jar contains 11 sweets. 7 of the sweets are green and 4 are yellow. Kofi takes a sweet at random and eats it. He then takes a second sweet at random and eats it.
(a)
The first sweet Kofi takes is green. Write down the probability that the second sweet is yellow.
(1)
(b)
Work out the probability that both sweets are the same colour.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 for 4/10 oe
✓ 1
Part (b)
M1 for 7/11 × 6/10 or 4/11 × 3/10
✓ 1
M1 for 7/11 × 6/10 + 4/11 × 3/10
✓ 1
A1 for 54/110 = 27/55 oe
✓ 1
Exam tip: When items are not replaced, both the numerator and the denominator change on the second branch of the tree. Check that each pair of branches still adds to 1.
20IGME-063-03-04Conditional probability and tree diagramsPaper HEasy5 marks
120 students each study exactly one of French or Spanish. In Year 10, 28 students study French and 22 study Spanish. In Year 11, 39 students study French and 31 study Spanish. One of the 120 students is chosen at random.
(a)
Find the probability that the student studies Spanish.
(1)
(b)
Given that the student is in Year 11, find the probability that the student studies French.
(2)
(c)
Given that the student studies French, find the probability that the student is in Year 10.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 for 53/120 oe
✓ 1
Part (b)
M1 for a fraction with denominator 70
✓ 1
A1 for 39/70 oe
✓ 1
Part (c)
M1 for a fraction with denominator 28 + 39 = 67
✓ 1
A1 for 28/67 oe
✓ 1
Common error: "Given that" restricts you to one group, so the denominator is the size of that group, not the total of 120.
21IGME-063-03-05Conditional probability and tree diagramsPaper HEasy5 marks
The probability that it rains on a school day is 0.3. If it rains, the probability that Ama walks to school is 0.2. If it does not rain, the probability that Ama walks to school is 0.65.
(a)
Work out the probability that on a school day it rains and Ama does not walk to school.
(2)
(b)
Work out the probability that Ama walks to school on a school day.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 for 0.3 × 0.8
✓ 1
A1 for 0.24
✓ 1
Part (b)
M1 for 0.3 × 0.2 or 0.7 × 0.65
✓ 1
M1 for 0.3 × 0.2 + 0.7 × 0.65
✓ 1
A1 for 0.515
✓ 1
Exam tip: Sketch a tree diagram even when one is not asked for: the second set of branches depends on the weather, so the walking probabilities differ on each branch.
22IGME-063-02-02Cumulative frequency and histogramsPaper HEasy4 marks
The masses, m kg, of 80 parcels are recorded.
0 < m ≤ 2 : frequency 12 2 < m ≤ 5 : frequency 27 5 < m ≤ 10 : frequency 25 10 < m ≤ 20 : frequency 16
The information is to be shown in a histogram.
(a)
Work out the frequency density for each class.
(2)
(b)
On the histogram, the bar for the class 2 < m ≤ 5 is 4.5 cm tall. Work out the height of the bar for the class 10 < m ≤ 20.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 frequency ÷ class width for at least two classes
✓ 1
A1 6, 9, 5, 1.6
✓ 1
Part (b)
M1 1.6 × (4.5 ÷ 9) or 0.5 cm per unit of frequency density
✓ 1
A1 0.8 cm
✓ 1
Common error: Plotting frequency as the bar height is wrong when class widths differ; the height must be frequency density.
23IGME-063-02-03Cumulative frequency and histogramsPaper HEasy3 marks
A histogram shows the heights, h cm, of some seedlings. The frequency densities of the bars are:
0 < h ≤ 4 : 2.5 4 < h ≤ 6 : 8 6 < h ≤ 10 : 4.5 10 < h ≤ 15 : 1.2
(a)
Work out the frequency for each class.
(2)
(b)
Work out the total number of seedlings.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 frequency density × class width for at least two classes
✓ 1
A1 10, 16, 18, 6
✓ 1
Part (b)
B1 50
✓ 1
Exam tip: In a histogram the area of a bar represents frequency, so multiply the frequency density by the class width.
24IGME-063-02-05Cumulative frequency and histogramsPaper HEasy3 marks
The lengths, L minutes, of 200 phone calls to a helpline are summarised.
L ≤ 2 : 28 L ≤ 4 : 76 L ≤ 6 : 132 L ≤ 8 : 170 L ≤ 10 : 188 L ≤ 12 : 200
(a)
How many calls lasted longer than 6 minutes?
(1)
(b)
Work out the percentage of calls that lasted more than 4 minutes but at most 8 minutes.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 200 − 132 = 68
✓ 1
Part (b)
M1 (170 − 76) ÷ 200 or 94/200
✓ 1
A1 47%
✓ 1
Common error: Reading the cumulative frequency at 6 minutes gives the number of calls up to 6 minutes; subtract from the total for 'longer than'.
25IGME-063-02-01Cumulative frequency and histogramsPaper HEasy5 marks
The table gives information about the times, t minutes, that 50 customers spent waiting at a bank.
0 < t ≤ 10 : frequency 4 10 < t ≤ 20 : frequency 11 20 < t ≤ 30 : frequency 18 30 < t ≤ 40 : frequency 12 40 < t ≤ 50 : frequency 5
(a)
Write down the cumulative frequencies for t ≤ 10, t ≤ 20, t ≤ 30, t ≤ 40 and t ≤ 50.
(2)
(b)
Write down the number of customers who waited more than 30 minutes.
(1)
(c)
Assume the times are spread evenly within each class. Use linear interpolation to estimate the median waiting time.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B2 for 4, 15, 33, 45, 50
✓ 2
(B1 for at least 3 correct values)
—
Part (b)
B1 50 − 33 = 17
✓ 1
Part (c)
M1 the median (25th value) lies in the class 20 < t ≤ 30, so 20 + (25 − 15)/18 × 10
✓ 1
A1 25.6 minutes (accept 25.5 to 25.8, which also allows for using the 25.5th value)
✓ 1
Exam tip: When you interpolate, start from the lower boundary of the median class. Use the cumulative frequency up to that boundary, not the class frequency.
26IGME-063-01-05Probability of single and combined eventsPaper HEasy5 marks
The probability that Amira's train is late on any day is 0.12. Whether the train is late on one day is independent of any other day.
(a)
Write down the probability that the train is not late on a given day.
(1)
(b)
Work out the probability that the train is late on both Monday and Tuesday.
(2)
(c)
Work out the probability that the train is late on exactly one of the two days, Monday and Tuesday.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1: 0.88
✓ 1
Part (b)
M1: 0.12 × 0.12
✓ 1
A1: 0.0144
✓ 1
Part (c)
M1: 0.12 × 0.88 + 0.88 × 0.12, or 2 × 0.12 × 0.88
✓ 1
A1: 0.2112
✓ 1
Common error: 'Exactly one' can happen in two ways: late then on time, or on time then late. Add both products. A single product of 0.1056 is not enough.
27IGME-063-00-09Averages, range and statistical diagramsPaper HMedium7 marks
The heights, h cm, of 80 plants are summarised below. 0 < h ≤ 10: 6 10 < h ≤ 20: 14 20 < h ≤ 30: 25 30 < h ≤ 40: 22 40 < h ≤ 50: 13
(a)
Work out the cumulative frequency for each class.
(2)
(b)
Write down the class interval that contains the upper quartile.
(1)
(c)
Work out the percentage of the plants that are taller than 30 cm.
(2)
(d)
Assuming the heights are spread evenly within each class, estimate the median height.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 for at least three correct cumulative frequencies
✓ 1
B1 for all correct: 6, 20, 45, 67, 80
✓ 1
Part (b)
B1 for 30 < h ≤ 40, since the 60th value lies between cumulative frequencies 45 and 67
✓ 1
Part (c)
M1 for (22 + 13) ÷ 80, or 35/80
✓ 1
A1 for 43.75% (accept 43.8%)
✓ 1
Part (d)
M1 for identifying the 40th value in the class 20 < h ≤ 30 and using 20 + (40 − 20)/25 × 10
✓ 1
A1 for 28 cm
✓ 1
Exam tip: For a large grouped data set, use n ÷ 2 = 40 for the median position and 3n ÷ 4 = 60 for the upper quartile. Cumulative frequencies show which class each value lies in.
28IGME-063-00-10Averages, range and statistical diagramsPaper HMedium6 marks
A histogram shows the commuting times, t minutes, of a group of workers. The frequency densities of the bars are: 0 < t ≤ 5: 2.4 5 < t ≤ 10: 4.0 10 < t ≤ 20: 3.2 20 < t ≤ 40: 0.8
(a)
Work out the total number of workers in the group.
(3)
(b)
Work out an estimate for the number of workers whose commuting time was more than 15 minutes.
(2)
(c)
Express your answer to part (b) as a percentage of all the workers.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 for using frequency = frequency density × class width
✓ 1
M1 for at least three correct frequencies from 12, 20, 32, 16
✓ 1
A1 for 80 workers
✓ 1
Part (b)
M1 for 5 × 3.2 = 16 from the 10 < t ≤ 20 class, added to 16 from the 20 < t ≤ 40 class
✓ 1
A1 for 32 workers
✓ 1
Part (c)
B1 for 32 ÷ 80 × 100 = 40%
✓ 1
Common error: The height of a histogram bar is frequency density, not frequency. Multiply by the class width to find each frequency.
29IGME-063-01-06Probability of single and combined eventsPaper HMedium5 marks
A box contains 6 red pens and 4 green pens only. Mo takes two pens at random from the box without replacement.
(a)
Work out the probability that both pens are green.
(2)
(b)
Work out the probability that the two pens are different colours.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: \(\frac{4}{10} \times \frac{3}{9}\)
✓ 1
A1: \(\frac{12}{90}\) or \(\frac{2}{15}\)
✓ 1
Part (b)
M1: \(\frac{6}{10} \times \frac{4}{9}\) or \(\frac{4}{10} \times \frac{6}{9}\)
✓ 1
M1: \(\frac{24}{90} + \frac{24}{90}\)
✓ 1
A1: \(\frac{48}{90}\) or \(\frac{8}{15}\)
✓ 1
Common error: Without replacement, both the number of that colour and the total go down by 1 for the second pick.
30IGME-063-01-09Probability of single and combined eventsPaper HMedium5 marks
A biased four-sided dice is numbered 1, 2, 3 and 4. P(1) = 0.2, P(2) = x, P(3) = 2x, P(4) = 0.35
(a)
Work out the value of x.
(2)
(b)
The dice is rolled twice. Work out the probability that the total of the two scores is 4.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: 0.2 + x + 2x + 0.35 = 1
✓ 1
A1: x = 0.15
✓ 1
Part (b)
M1: identifies (1,3), (3,1) and (2,2) as the only ways
✓ 1
M1: 0.2 × 0.3 + 0.3 × 0.2 + 0.15 × 0.15
✓ 1
A1: 0.1425
✓ 1
Common error: (1,3) and (3,1) are different outcomes, so each must be counted, but (2,2) is counted only once.
31IGME-063-01-10Probability of single and combined eventsPaper HMedium5 marks
Each time Jade takes a basketball free throw, the probability that she scores is 0.7, independently of any other throw. Jade takes 3 free throws.
(a)
Work out the probability that she scores at least once.
(2)
(b)
Work out the probability that she scores exactly twice.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: 1 − 0.3 × 0.3 × 0.3
✓ 1
A1: 0.973
✓ 1
Part (b)
M1: 0.7 × 0.7 × 0.3
✓ 1
M1: multiplies by 3 for the three possible orders
✓ 1
A1: 0.441
✓ 1
Exam tip: For 'at least one', it is usually quickest to work out 1 minus the probability of none.
32IGME-063-01-11Probability of single and combined eventsPaper HMedium4 marks
There are 30 students in a club. 17 play football, 12 play tennis and 6 play both football and tennis.
(a)
A student is chosen at random. Work out the probability that the student plays neither sport.
(2)
(b)
A student who plays tennis is chosen at random. Work out the probability that this student also plays football.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: 17 + 12 − 6 = 23, so 30 − 23 = 7 play neither
✓ 1
A1: \(\frac{7}{30}\)
✓ 1
Part (b)
M1: \(\frac{6}{12}\)
✓ 1
A1: \(\frac{1}{2}\)
✓ 1
Common error: Adding 17 and 12 counts the 6 students who play both twice. Sketch a Venn diagram and fill in the overlap first.
33IGME-063-03-08Conditional probability and tree diagramsPaper HMedium7 marks
A bag contains x counters. 3 of the counters are red and the rest are green. Two counters are taken at random from the bag without replacement. The probability that both counters are red is 1/15.
(a)
Show that \(x^2 - x - 90 = 0\)
(3)
(b)
Solve \(x^2 - x - 90 = 0\) to find the number of counters in the bag.
(2)
(c)
Work out the probability that both counters taken are green.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 for 3/x × 2/(x − 1)
✓ 1
M1 for 6/(x(x − 1)) = 1/15 leading to x(x − 1) = 90
✓ 1
A1 for correct rearrangement to \(x^2 - x - 90 = 0\) with no errors
✓ 1
Part (b)
M1 for (x − 10)(x + 9) = 0 or correct use of the quadratic formula
✓ 1
A1 for x = 10, with −9 rejected
✓ 1
Part (c)
M1 for 7/10 × 6/9
✓ 1
A1 for 7/15 oe
✓ 1
Exam tip: In a "show that" part every step must be visible, including clearing the fractions and collecting terms on one side.
34IGME-063-03-09Conditional probability and tree diagramsPaper HMedium6 marks
A sports club has 50 members. 32 members play tennis. 25 members play squash. 6 members play neither tennis nor squash.
(a)
Work out how many members play both tennis and squash.
(2)
(b)
A member who plays squash is chosen at random. Find the probability that this member also plays tennis.
(2)
(c)
A member who plays exactly one of the two sports is chosen at random. Find the probability that this member plays tennis.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 for 32 + 25 − (50 − 6) oe
✓ 1
A1 for 13
✓ 1
Part (b)
M1 for a fraction with denominator 25
✓ 1
A1 for 13/25 oe
✓ 1
Part (c)
M1 for identifying 19 (tennis only) and 12 (squash only)
✓ 1
A1 for 19/31 oe
✓ 1
Common error: "Exactly one sport" excludes the 13 members who play both, so the denominator is 19 + 12, not 32 + 25.
35IGME-063-03-10Conditional probability and tree diagramsPaper HMedium6 marks
The probability that Sam hits a target with each arrow is 0.7, independently of other arrows. Sam keeps shooting until he hits the target, but he is allowed a maximum of three arrows.
(a)
Work out the probability that Sam hits the target with his third arrow.
(2)
(b)
Work out the probability that Sam hits the target.
(2)
(c)
Given that Sam hits the target, find the probability that he hit it with his first arrow. Give your answer correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 for 0.3 × 0.3 × 0.7
✓ 1
A1 for 0.063
✓ 1
Part (b)
M1 for 1 − 0.33 or 0.7 + 0.3 × 0.7 + 0.063
✓ 1
A1 for 0.973
✓ 1
Part (c)
M1 for 0.7 ÷ 0.973
✓ 1
A1 for 0.719
✓ 1
Common error: The answer to the last part is not simply 0.7: the condition "he hits the target" removes the outcome where all three arrows miss.
36IGME-063-03-11Conditional probability and tree diagramsPaper HMedium5 marks
The probability that Dev passes his driving test at the first attempt is 0.55. If he fails, the probability that he passes at the second attempt is 0.7. If he fails the second attempt, the probability that he passes at the third attempt is 0.8.
(a)
Work out the probability that Dev passes his test at either the first or the second attempt.
(3)
(b)
Work out the probability that Dev fails at all of his first three attempts.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 for 0.45 × 0.7
✓ 1
M1 for 0.55 + 0.45 × 0.7
✓ 1
A1 for 0.865
✓ 1
Part (b)
M1 for 0.45 × 0.3 × 0.2
✓ 1
A1 for 0.027
✓ 1
Exam tip: Each new attempt only happens on the "fail" branch, so the tree only extends from the fail branches.
37IGME-063-02-06Cumulative frequency and histogramsPaper HMedium5 marks
A sports club has 150 members. Their ages, a years, are grouped.
10 < a ≤ 20 : frequency not given 20 < a ≤ 30 : frequency 36 30 < a ≤ 50 : frequency not given 50 < a ≤ 80 : frequency 27
In a histogram of the data, the frequency density of the class 10 < a ≤ 20 is 2.4.
(a)
Find the number of members aged 10 < a ≤ 20.
(2)
(b)
Work out the frequency density for the class 30 < a ≤ 50.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 2.4 × 10
✓ 1
A1 24
✓ 1
Part (b)
M1 150 − (24 + 36 + 27)
✓ 1
A1 frequency = 63
✓ 1
A1 63 ÷ 20 = 3.15
✓ 1
Exam tip: Use the total to find a missing frequency, then divide by the class width (20, not 10) for the density.
38IGME-063-02-08Cumulative frequency and histogramsPaper HMedium5 marks
A histogram shows the masses, m grams, of apples from an orchard. The frequency densities are:
80 < m ≤ 100 : 0.6 100 < m ≤ 110 : 2.4 110 < m ≤ 120 : 3.0 120 < m ≤ 150 : 0.5
(a)
Work out the frequency for each class.
(2)
(b)
Work out an estimate for the mean mass of the apples.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 frequency density × class width for at least two classes
Common error: Dividing by the number of classes instead of the total frequency (81) gives a far too large mean.
39IGME-063-02-10Cumulative frequency and histogramsPaper HMedium6 marks
The lifetimes, h hours, of 200 batteries of brand X were tested.
h ≤ 30 : 10 h ≤ 35 : 50 h ≤ 40 : 110 h ≤ 45 : 170 h ≤ 50 : 190 h ≤ 60 : 200
Assume the cumulative frequency graph is made of straight lines joining the plotted points. For brand Y, the median lifetime is 42 hours and the interquartile range is 15 hours.
(a)
Estimate the median lifetime of brand X batteries.
(2)
(b)
Estimate the interquartile range of the lifetimes of brand X batteries.
B1 brand Y has a higher median so its batteries last longer on average
✓ 1
B1 brand X has a smaller IQR so its lifetimes are more consistent
✓ 1
Exam tip: When comparing, make one comment about an average and one about spread, and relate each to the context.
40IGME-063-02-11Cumulative frequency and histogramsPaper HMedium5 marks
A histogram shows the times, t minutes, that students spent on homework one evening. No scale is given on the frequency density axis.
The bar for 0 < t ≤ 5 is 3.6 cm tall and represents 27 students. The bar for 5 < t ≤ 15 is 2.4 cm tall. The bar for 15 < t ≤ 30 is 0.8 cm tall.
(a)
Work out the number of students who spent between 5 and 15 minutes on homework.
(3)
(b)
Work out the number of students who spent between 15 and 30 minutes on homework.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 frequency density of first bar = 27 ÷ 5 = 5.4
✓ 1
M1 scale: 5.4 ÷ 3.6 = 1.5 units per cm, so second bar density = 3.6
✓ 1
A1 3.6 × 10 = 36
✓ 1
Part (b)
M1 0.8 × 1.5 × 15
✓ 1
A1 18
✓ 1
Common error: Assuming 1 cm represents 1 unit of frequency density without checking against the known bar gives wrong frequencies.
41IGME-063-03-06Conditional probability and tree diagramsPaper HMedium5 marks
A box contains 10 balls. 6 of the balls are white and 4 are black. Three balls are taken at random from the box, one after another, without replacement.
(a)
Work out the probability that all three balls are black.
(2)
(b)
Work out the probability that exactly two of the balls are white.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 for 4/10 × 3/9 × 2/8
✓ 1
A1 for 1/30 oe
✓ 1
Part (b)
M1 for the probability of one order of two white and one black, e.g. P(W, B, W) = 6/10 × 4/9 × 5/8 or P(W, W, B) = 6/10 × 5/9 × 4/8 (black third), each = 1/6
✓ 1
M1 for 3 × 1/6, or the sum of the three orders WWB + WBW + BWW. Alternatively, award both method marks for (⁶C₂ × ⁴C₁) ÷ ¹⁰C₃ = (15 × 4) ÷ 120
✓ 1
A1 for 1/2 oe
✓ 1
Common error: Two white and one black can occur in three orders (WWB, WBW, BWW). Each order has the same probability, 1/6, so multiply one product by 3. Check that one factor is a black probability, so that you have not worked out P(three white) by mistake.
42IGME-063-02-07Cumulative frequency and histogramsPaper HMedium6 marks
The speeds, v mph, of cars passing a school were recorded. A histogram of the data has these frequency densities:
20 < v ≤ 30 : 1.2 30 < v ≤ 35 : 6.4 35 < v ≤ 40 : 8.8 40 < v ≤ 50 : 3.2 50 < v ≤ 70 : 0.4
(a)
Work out the total number of cars.
(2)
(b)
Estimate the number of cars with a speed greater than 45 mph.
(2)
(c)
The speed limit is 40 mph. Work out the percentage of cars that were exceeding the speed limit.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 12 + 32 + 44 + 32 + 8 (at least three frequencies correct)
✓ 1
A1 128
✓ 1
Part (b)
M1 3.2 × 5 (= 16) + 8
✓ 1
A1 24
✓ 1
Part (c)
M1 (32 + 8) ÷ 128
✓ 1
A1 31.25% (accept 31.3%)
✓ 1
Exam tip: For part of a class, assume the values are spread evenly: half the class width gives half the frequency.
43IGME-063-03-07Conditional probability and tree diagramsPaper HMedium5 marks
A factory has two machines, A and B, that make light bulbs. Machine A makes 60% of the bulbs and machine B makes the rest. 3% of the bulbs made by machine A are faulty. 5% of the bulbs made by machine B are faulty. A bulb is chosen at random.
(a)
Work out the probability that the bulb is faulty.
(3)
(b)
Given that the bulb is faulty, find the probability that it was made by machine B. Give your answer correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 for 0.6 × 0.03 (= 0.018) or 0.4 × 0.05 (= 0.02)
✓ 1
M1 for 0.6 × 0.03 + 0.4 × 0.05
✓ 1
A1 for 0.038
✓ 1
Part (b)
M1 for (0.4 × 0.05) ÷ 0.038 or 0.02 ÷ 0.038
✓ 1
A1 for 0.526
✓ 1
Exam tip: For a reversed conditional probability, divide the probability of the branch you want by the total probability of the given event.
44IGME-063-02-09Cumulative frequency and histogramsPaper HMedium5 marks
160 students sat a test marked out of 100. The cumulative frequency table shows their marks, x.
x ≤ 20 : 8 x ≤ 40 : 30 x ≤ 50 : 62 x ≤ 60 : 110 x ≤ 70 : 140 x ≤ 80 : 152 x ≤ 100 : 160
Assume the cumulative frequency graph is made of straight lines joining the plotted points.
(a)
Estimate the median mark.
(2)
(b)
The top 10% of students are awarded a distinction. Estimate the lowest mark needed for a distinction.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 the 80th value lies between 50 and 60: 50 + (80 − 62)/48 × 10 (or uses 80.5: 50 + (80.5 − 62)/48 × 10)
✓ 1
A1 53.75 (accept any value from 53.7 to 53.9)
✓ 1
Part (b)
M1 10% of 160 = 16, so cumulative frequency 160 − 16 = 144
✓ 1
M1 70 + (144 − 140)/12 × 10
✓ 1
A1 73.3 (accept any value from 73 to 73.4)
✓ 1
Common error: Reading off at cumulative frequency 16 finds the bottom 10%, not the top 10%. Subtract from the total first.
45IGME-063-00-14Averages, range and statistical diagramsPaper HHard6 marks
A group of students contains g girls and b boys. The mean mass of the girls is 52 kg. The mean mass of the boys is 60 kg. The mean mass of the whole group is 55.2 kg.
(a)
Show that g : b = 3 : 2
(4)
(b)
There are 45 students in the group. Work out the number of girls.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 for total mass = 52g + 60b
✓ 1
M1 for 52g + 60b = 55.2(g + b)
✓ 1
M1 for 4.8b = 3.2g
✓ 1
A1 for g/b = 4.8/3.2 = 3/2, so g : b = 3 : 2 shown
✓ 1
Part (b)
M1 for 45 ÷ (3 + 2) × 3
✓ 1
A1 for 27 girls
✓ 1
Common error: The combined mean is not 56, the average of 52 and 60. It is closer to 52, so there must be more girls than boys, which is a good check on your ratio.
46IGME-063-01-12Probability of single and combined eventsPaper HHard7 marks
A bag contains n counters. 3 of the counters are red and the rest are blue. Two counters are taken at random from the bag without replacement. The probability that both counters are red is \(\frac{1}{15}\).
(a)
Show that \(n^2 - n - 90 = 0\).
(3)
(b)
Solve the equation to find the value of n.
(2)
(c)
Work out the probability that exactly one of the two counters is red.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: \(\frac{3}{n} \times \frac{2}{n-1}\)
✓ 1
M1: \(\frac{6}{n(n-1)} = \frac{1}{15}\) leading to \(n(n-1) = 90\)
✓ 1
A1: correctly reaches \(n^2 - n - 90 = 0\)
✓ 1
Part (b)
M1: \((n-10)(n+9) = 0\) or correct use of the quadratic formula
Exam tip: In a 'show that' question, every algebraic step must be written down, ending with exactly the given equation.
47IGME-063-01-13Probability of single and combined eventsPaper HHard5 marks
The probability that it rains on a given morning is 0.35. If it rains, the probability that Tom is late for work is 0.4. If it does not rain, the probability that Tom is late for work is 0.1.
(a)
Work out the probability that Tom is late for work on a given morning.
(2)
(b)
Tom is late for work one morning. Work out the probability that it rained that morning. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: 0.35 × 0.4 + 0.65 × 0.1
✓ 1
A1: 0.205
✓ 1
Part (b)
M1: 0.35 × 0.4 = 0.14 as the numerator
✓ 1
M1: 0.14 ÷ 0.205
✓ 1
A1: 0.683
✓ 1
Common error: The answer to the second part is not 0.4. Knowing Tom was late restricts the outcomes, so divide by the probability of being late.
48IGME-063-03-12Conditional probability and tree diagramsPaper HHard5 marks
Bag A contains 3 red counters and 5 blue counters. Bag B contains 4 red counters and 2 blue counters. A counter is taken at random from bag A and placed in bag B. A counter is then taken at random from bag B.
(a)
Show that the probability that the counter taken from bag B is red is 5/8
(3)
(b)
Given that the counter taken from bag B is red, find the probability that the counter moved from bag A was red.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 for 3/8 × 5/7 or 5/8 × 4/7
✓ 1
M1 for 3/8 × 5/7 + 5/8 × 4/7
✓ 1
A1 for 15/56 + 20/56 = 35/56 = 5/8 with working shown
✓ 1
Part (b)
M1 for (15/56) ÷ (35/56)
✓ 1
A1 for 3/7 oe
✓ 1
Common error: After the transfer, bag B contains 7 counters, not 6. The number of red counters in bag B also depends on which colour was moved.
49IGME-063-03-14Conditional probability and tree diagramsPaper HHard5 marks
A box contains 12 pens: 5 black, 4 blue and 3 red. Ben takes two pens at random from the box without replacement.
(a)
Work out the probability that the two pens are the same colour.
(3)
(b)
Given that the two pens are the same colour, find the probability that they are both black.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 for one of 5/12 × 4/11, 4/12 × 3/11 or 3/12 × 2/11
✓ 1
M1 for 5/12 × 4/11 + 4/12 × 3/11 + 3/12 × 2/11
✓ 1
A1 for 38/132 = 19/66 oe
✓ 1
Part (b)
M1 for (20/132) ÷ (38/132)
✓ 1
A1 for 10/19 oe
✓ 1
Common error: Giving 20/132 for the last part ignores the condition. Divide by the probability that the pens are the same colour.
50IGME-063-02-12Cumulative frequency and histogramsPaper HHard5 marks
A histogram shows the lengths, l cm, of fish caught in a lake. The frequency densities are:
20 < l ≤ 30 : 1.5 30 < l ≤ 35 : 5.2 35 < l ≤ 40 : 6.0 40 < l ≤ 50 : 1.6 50 < l ≤ 60 : 0.3
(a)
Work out the total number of fish.
(2)
(b)
Two of the fish are chosen at random without replacement. Work out the probability that both fish are more than 40 cm long.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 15 + 26 + 30 + 16 + 3 (at least three frequencies correct)
✓ 1
A1 90
✓ 1
Part (b)
M1 16 + 3 = 19 fish more than 40 cm
✓ 1
M1 19/90 × 18/89
✓ 1
A1 19/445 or 0.0427
✓ 1
Common error: Using 19/90 × 19/90 ignores that the first fish is not replaced, so both numerator and denominator drop by 1.
51IGME-063-02-13Cumulative frequency and histogramsPaper HHard5 marks
80 people were asked how long, t minutes, they spent travelling to work. The data are shown in a histogram.
• The class 0 < t ≤ 10 has frequency density 1.2 • The class 10 < t ≤ 15 has frequency density three times that of the class 0 < t ≤ 10 • The class 15 < t ≤ 25 has frequency x • The class 25 < t ≤ 40 has frequency density 0.4
(a)
Find the value of x.
(3)
(b)
Estimate the number of people who spent more than 20 minutes travelling to work.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 frequencies 1.2 × 10 = 12 or 3.6 × 5 = 18 or 0.4 × 15 = 6
✓ 1
M1 80 − (12 + 18 + 6)
✓ 1
A1 x = 44
✓ 1
Part (b)
M1 44 ÷ 2 (= 22) + 6
✓ 1
A1 28
✓ 1
Exam tip: Turn every frequency density into a frequency first; the total then gives the unknown directly.
52IGME-063-02-14Cumulative frequency and histogramsPaper HHard6 marks
A histogram shows the waiting times, w minutes, of patients at a clinic. The frequency densities are:
0 < w ≤ 4 : 2.5 4 < w ≤ 6 : 9 6 < w ≤ 10 : 6.5 10 < w ≤ 20 : 0.6
Assume the values within each class are evenly spread.
(a)
Work out an estimate for the median waiting time.
(3)
(b)
Work out an estimate for the interquartile range of the waiting times.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 frequencies 10, 18, 26, 6 (total 60)
✓ 1
M1 30th value: 6 + (30 − 28)/26 × 4
✓ 1
A1 6.31 minutes
✓ 1
Part (b)
M1 LQ (15th value) = 4 + (15 − 10)/18 × 2 = 4.56
✓ 1
M1 UQ (45th value) = 6 + (45 − 28)/26 × 4 = 8.62
✓ 1
A1 4.06 minutes (accept 4.05–4.07)
✓ 1
Exam tip: Build a running total of frequencies to locate which class each quartile is in before interpolating.
53IGME-063-01-14Probability of single and combined eventsPaper HHard8 marks
A jar contains 5 red sweets, 4 blue sweets and 3 yellow sweets only. Priya takes three sweets at random from the jar without replacement.
(a)
Work out the probability that all three sweets are red.
(2)
(b)
Work out the probability that all three sweets are the same colour.
(3)
(c)
Work out the probability that Priya takes one sweet of each colour.
M1: all blue \(\frac{4}{12} \times \frac{3}{11} \times \frac{2}{10} = \frac{24}{1320}\), or all yellow \(\frac{3}{12} \times \frac{2}{11} \times \frac{1}{10} = \frac{6}{1320}\)
✓ 1
M1: \(\frac{60}{1320} + \frac{24}{1320} + \frac{6}{1320}\), adding all three single-colour probabilities
✓ 1
A1: \(\frac{90}{1320}\) or \(\frac{3}{44}\)
✓ 1
Part (c)
M1: P(red, then blue, then yellow) \(= \frac{5}{12} \times \frac{4}{11} \times \frac{3}{10} = \frac{60}{1320}\) for one order
✓ 1
M1: multiplies by 6 for the different orders
✓ 1
A1: \(\frac{360}{1320}\) or \(\frac{3}{11}\)
✓ 1
Alternative: M1 for 5 × 4 × 3 = 60 selections with one of each colour; M1 for dividing by \(^{12}C_3 = 220\); A1 for \(\frac{3}{11}\)
—
Exam tip: Three different colours can be taken in 3 × 2 × 1 = 6 orders, and each order has the same probability.
54IGME-063-03-13Conditional probability and tree diagramsPaper HHard6 marks
A biased spinner lands on red with probability p, where p < 0.5. The spinner is spun twice. The probability that it lands on red exactly once is 0.42.
(a)
Show that \(100p^2 - 100p + 21 = 0\)
(2)
(b)
Find the value of p.
(2)
(c)
The spinner is spun three times. Work out the probability that it lands on red at least once.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 for 2p(1 − p) = 0.42 or p(1 − p) + (1 − p)p = 0.42
✓ 1
A1 for a correct intermediate step, e.g. 2p − 2p2 = 0.42 or 100p − 100p2 = 21, leading to \(100p^2 - 100p + 21 = 0\) with no errors
✓ 1
Part (b)
M1 for (10p − 3)(10p − 7) = 0 or correct use of the quadratic formula
✓ 1
A1 for p = 0.3, with 0.7 rejected because p < 0.5
✓ 1
Part (c)
M1 for 1 − 0.73 or equivalent full method
✓ 1
A1 for 0.657
✓ 1
Exam tip: "Exactly once" means red then not red, or not red then red. The probability is therefore 2p(1 − p), not p(1 − p). In a 'show that' question, write down every step of the rearrangement.
55IGME-063-00-12Averages, range and statistical diagramsPaper HHard8 marks
The ages, a years, of 120 people at a concert are summarised below. 0 < a ≤ 10: frequency 15 10 < a ≤ 25: frequency unknown; the frequency density of this class on a histogram is 2.4 25 < a ≤ 40: frequency 30 40 < a ≤ 70: frequency unknown
(a)
Work out the number of people aged 10 < a ≤ 25.
(2)
(b)
Work out the frequency density of the bar for the class 40 < a ≤ 70.
(3)
(c)
Assuming the ages are spread evenly within each class, work out an estimate for the median age.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 for 2.4 × 15
✓ 1
A1 for 36 people
✓ 1
Part (b)
M1 for 120 − 15 − 36 − 30
✓ 1
A1 for a frequency of 39
✓ 1
A1 for frequency density 39 ÷ 30 = 1.3
✓ 1
Part (c)
M1 for the 60th value (or the 60.5th value), with cumulative frequencies 15, 51, 81 placing it in 25 < a ≤ 40
✓ 1
M1 for 25 + (60 − 51)/30 × 15, or 25 + (60.5 − 51)/30 × 15
✓ 1
A1 for 29.5 years (accept 29.75 years from using the 60.5th value)
✓ 1
Exam tip: Check the class widths carefully: here they are 10, 15, 15 and 30. Frequency = frequency density × class width, so using the wrong width gives the wrong frequency in every later part.
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